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Chapter 31 Nuclear Physics and Radioactivity
Chapter 31
NUCLEAR PHYSICS AND RADIOACTIVITY
PREVIEW
The modern view of the atom includes electrons in energy levels around the nucleus of
the atom. The nucleus contains positively-charged protons and neutral neutrons, each of
which are made up of quarks. The nucleus is held together by the strong nuclear force,
and the binding energy in the nucleus is a result of some of the mass of the particles (the
mass defect) in the nucleus being converted into energy by the relationship E = mc2. The
atomic number is the number of protons in the nucleus, and the atomic mass number is
the number of nucleons (protons and neutrons) in the nucleus. Nuclear changes can take
place, but the total amount of atomic mass in the process must remain constant.
The content contained in sections 1 – 4, and 10 (Example 13) of chapter 31 of the
textbook is included on the AP Physics B exam.
QUICK REFERENCE
Important Terms
alpha particle
positively charged particle consisting of two protons and two neutrons
atomic mass number (A)
the number of protons and neutrons in the nucleus of an atom
atomic mass unit
the unit of mass equal to 1/12 the mass of a carbon-12 nucleus; the
atomic mass rounded to the nearest whole number is called the mass number
atomic number (Z)
the number of protons in the nucleus of an atom
beta particle
high speed electron emitted from a radioactive element when a neutron
decays into a proton
binding energy
the nuclear energy that binds protons and neutrons in the nucleus of the
atom
element
a substance made of only one kind of atom
isotope
a form of an element which has a particular number of neutrons, that is, has the
same atomic number but a different mass number than the other elements which
occupy the same place on the periodic table
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Chapter 31 Nuclear Physics and Radioactivity
mass defect
the mass equivalent of the binding energy in the nucleus of an atom by E =
mc2
neutron
an electrically neutral subatomic particle found in the nucleus of an atom
nuclear reaction
any process in the nucleus of an atom that causes the number of
protons and/or neutrons to change
nucleons
protons or a neutrons
strong nuclear force
the force that binds protons and neutrons together in the nucleus of
an atom
transmutation
the changing of one element into another by a loss of gain of one or more protons
Equations and Symbols
E  mc 2
where
1u  1.6726 x10 27 kg  931.5 MeV
A
Z X
ΔE = binding energy of the nucleus
Δm = mass defect of the nucleus
c = speed of light = 3 x 108 m/s
u = atomic mass unit
X = element symbol
A = atomic mass number (number of
protons and neutrons)
Z = atomic number (number of protons)
Ten Homework Problems
Chapter 31 Problems 1, 2, 10, 12, 14, 17, 18, 19, 22, 51
DISCUSSION OF SELECTED SECTIONS
31.3 The Mass Defect of the Nucleus and Nuclear Binding Energy
The nucleus is made up of positively charged protons and neutrons, which have no
charge. The proton has exactly the same charge as an electron, but is positive. The
neutron is actually made up of a proton and an electron bound together to create the
neutral particle. A proton is about 1800 times more massive than an electron, which
makes a neutron only very slightly more massive than a proton. We say that a proton has
a mass of approximately one atomic mass unit, u. The atomic number (Z) of an element is
equal to the number of protons found in an atom of that element, and fundamentally is an
indication of the charge on the nucleus of that element. All atoms of a given element have
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Chapter 31 Nuclear Physics and Radioactivity
the same atomic number. In other words, the number of protons an atom has defines what
kind of element it is. The total number of neutrons and protons in an atom is called the
mass number (A) of that element. The symbol ZA X is used to show both the atomic
number and the mass number of an X atom, where Z is the atomic number and A is the
mass number.
Even though the number of protons must be the same for all atoms of a particular
element, the number of neutrons, and thus the mass number, can be different. Atoms of
the same element with different masses are known as isotopes of one another. For
example, carbon-12 is a carbon atom with 6 protons and 6 neutrons, while carbon-14 is a
carbon atom with 6 protons and 8 neutrons. We would write these two isotopes of carbon
as 126C and 146C .
The table below summarizes the basic features of protons, neutrons and electrons. Notice
that we use an H to symbolize the proton, since the proton is a hydrogen nucleus.
Particle
proton
Symbol
1
1H
neutron
electron
Relative mass
1
Charge
+1
Location
nucleus
n
1
0
nucleus
e or e-
0
-1
electron orbitals
around the
nucleus
1
0
0
1
Example 1
Find the number of protons, electrons, and neutrons in a neutral atom of iron
56
26
Fe .
Solution
This isotope of iron has an atomic number of 26 and a mass number of 56. Therefore, it
will have 26 protons, 26 electrons, and 56 – 26 = 30 neutrons.
Since positive charges repel each other, one might wonder why protons stay together in
the nucleus of the atom. There is a force holding the protons together which is greater
than the electrostatic repulsion between them called the strong nuclear force, and it is a
result of the binding energy of the nucleus.
According to Einstein’s famous equation E = mc2, mass and energy can be converted into
one another. When an nucleus is assembled, each proton and neutron gives up a little of
its mass to be converted into binding energy.
Example 2
The nitrogen atom 147 N is composed of 7 protons and 7 neutrons, which gives a total of 14
atomic mass units (u). But if these particles are combined into a
the resulting mass of the nitrogen nucleus is 14.003074 u.
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14
7
N nitrogen nucleus,
Chapter 31 Nuclear Physics and Radioactivity
(a) Find the mass defect of the nitrogen nucleus.
(b) What is the binding energy of the nitrogen nucleus?
(c) What is the binding energy per nucleon?
Solution
(a) The sum of the masses of the protons and neutrons is
7(1.007 825 u) = 7.054775 u
7(1.008 665 u) = 7.060655 u
14.115430 u
The mass defect is
14.115430 u
- 14.003074 u
0.112356 u
(b) The equivalence between mass in atomic mass units and energy in million electronvolts (MeV) is 1 u = 931 MeV. Then the binding energy of the nucleus is
BE = (MD)(931 MeV/u) = (0.112356 u)(931 MeV/u) = 104.60344 MeV
(c)
104.60344 MeV
BE
MeV

 7.471674
nucleon
14 nucleons
nucleon
31.4 Radioactivity
At the end of the 19th century, there were elements discovered that continuously emitted
mysterious rays. These elements were identified as being radioactive. A radioactive
element spontaneously emits particles from its nucleus because the energy of the nucleus
is unstable. Examples of naturally-occurring radioactive elements are uranium 238
92 U ,
radium
226
88
Ra , and carbon 146C .
There are four types of particles that can be emitted when an element undergoes
radioactive decay:
1. Alpha decay . Uranium, for example, undergoes alpha decay, meaning that it emits an
alpha particle from its nucleus. An alpha particle is a helium nucleus, consisting of 2
protons and 2 neutrons. When an element emits an alpha particle, its nucleus loses 2
atomic numbers and 4 mass numbers, and thus changes into another element, called
the daughter element. But what would this element be? We can write the nuclear
equation for the radioactive decay of uranium as
U  Az X  24He
238
92
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Chapter 31 Nuclear Physics and Radioactivity
where X is the daughter element and 24 He is the alpha particle. The atomic number on the
left must equal the sum of the atomic numbers on the right, since charge and mass are
conserved in this process. The same is true for the mass numbers on the left and right. So,
the daughter element has an atomic number Z = 92-2 = 90 and a mass number A = 238 –
4 = 234. Uranium decays into the daughter element 234
90Th , thorium.
2. Beta decay. A beta particle is the name given to an electron emitted from the nucleus
of a radioactive element. But what is an electron doing in the nucleus of an atom?
Remember that we discussed the neutron in the nucleus of an atom as being a proton
and an electron bound together. Beta decay is really just a neutron emitting an
electron and becoming a proton. Thus, the daughter element resulting from beta decay
is one atomic number higher than the parent nucleus, but the mass number essentially
does not change. For example, carbon 146C is a radioactive element that undergoes
beta decay. The decay equation is
14
6
C  ZA X  10 e
We use the same symbol for a beta particle as we do for an electron. The daughter
element must have an atomic number of 6 – (-1) = 7 and a mass number of 14 – 0 = 14.
The daughter element is 147 N , nitrogen.
3. Gamma decay. Some radioactive elements emit a gamma ray, a very high energy
electromagnetic wave which has no charge or mass, so only the energy of the nucleus
changes, and neither Z nor A change.
4. Positron decay. A positron is exactly like an electron except for the fact that it is
positively charged. A positron is not a proton, as their masses and other features are
very different. Positron decay equations are typically not included on the AP Physics
B exam.
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Chapter 31 Nuclear Physics and Radioactivity
CHAPTER 31 REVIEW QUESTIONS
For each of the multiple choice questions below, choose the best answer.
1. The neutral element
24
magnesium 12
Mg has
(A) 12 protons, 12 electrons, and 24
neutrons.
(B) 12 protons, 12 electrons, and 12
neutrons.
(C) 24 protons, 24 electrons, and 12
neutrons.
(D) 24 protons, 12 electrons, and 12
neutrons.
(E) 12 protons, 24 electrons, and 24
neutrons.
4. The isotope of thorium 234
90Th
undergoes alpha decay according to the
equation
234
A
4
90Th  Z X  2 He . The element X is
2. All isotopes of uranium have
(A) the same atomic number and the
same mass number.
(B) different atomic numbers but the
same mass number.
(C) different atomic numbers and
different mass numbers.
(D) the same atomic number but
different mass numbers.
(E) no electrons.
(A)
238
92
(B)
230
88
Ra
(C)
236
94
Pu
(D)
238
88
Ra
(E)
230
92
U
U
60
Co undergoes
5. The isotope of cobalt 27
beta decay according to the equation
60
A
0
27 Co Z X  1 e . The element X is
60
Fe
(A) 26
(B)
56
25
Mn
(C)
60
28
Ni
60
29
(D) Cu
(E)
3. Six protons and six neutrons are
brought together to form a carbon
nucleus, but the mass of the carbon
nucleus is less than the sum of the
masses of the individual particles that
make up the nucleus. This missing mass,
called the mass defect, has been
(A) converted into the binding energy of
the nucleus.
(B) given off in a radioactive decay
process.
(C) converted into electrons.
(D) converted into energy to hold the
electrons in orbit.
(E) emitted as light.
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61
27
Co
Chapter 31 Nuclear Physics and Radioactivity
Free Response Question
Directions: Show all work in working the following question. The question is worth 10 points,
and the suggested time for answering the question is about 10 minutes. The parts within a
question may not have equal weight.
1. (10 points)
Particle or
nucleus
1
0n
Mass (u)
1.008 665 u
1
1
13
7
H
1.007 825 u
N
13.005738 u
14
7
N
14.003074 u
13
7
N  01n147N  energy
A neutron is bound to a nitrogen nucleus, as shown in the equation above.
(a) Find the mass defect and binding energy that holds a neutron to the nitrogen atom
(b) The U.S. uses about 1020 J of energy per year. How many nitrogen atoms would have to give
up a neutron and release its binding energy to provide the energy needed in the U.S. for a year?
ANSWERS AND EXPLANATIONS TO CHAPTER 31 REVIEW QUESTIONS
Multiple Choice
1. B
The atomic number 12 implies both 12 protons and 12 electrons, and the mass number 24 is the
sum of protons and neutrons, giving 12 neutrons.
2. D
All isotopes of a particular element must have the same atomic number (number of protons),
since this number identifies the element, but can have a different mass number (number of
neutrons).
3. A
Each particle that makes up the nucleus gives up a little mass to be converted into energy by E =
mc2 to bind the nucleus together.
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Chapter 31 Nuclear Physics and Radioactivity
4. B
The atomic number Z of element X is found by 90 = Z + 2, so Z = 88, and the mass number A is
found by 234 = A + 4, so A = 230. The element X is radium.
5. C
The atomic number Z of element X is found by 27 = Z + (-1), so Z = 28, and the mass number A
is found by 60 = A + 0, so A = 60. The element X is nickel.
Free Response Question Solution
(a) 5 points
13
7
N  01n147N  energy
(13.005738 u) + (1.008665 u) = 14.003074 u + (mass defect)
Mass defect = 0.011329 u
BE = (MD)(931 MeV/u) = (0.011329 u)( 931 MeV/u) = 10.547299 MeV
(b) 5 points
Converting MeV to J:
10.547299 MeV(1.6 x 10-13 J) = 1.69 x 10-12 J
Number of nitrogen atoms =
10 20 J
 5.93 x 10 31 atoms
12
1.69 x10 J / atom
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